MA1021 Calculus I B Term, 2014 Final Exam Print Name: Sign: Write up your solutions neatly and show all your work. 1. (28 pts) Compute each of the following derivatives: You do not have to simplify your results, but they must be neat, clear and correctly written up. a) d ( 5x 3 3x 5 + πx π 1 + πe π 1) = 5 3x 2 3 5x 4 + π(π 1)x π 2 + 0 (just the power rule.) b) d [ (1 + x 2 + x 4 + x 6 )(x + x 3 + x 5 + x 7 ) ] = (1 + x 2 + x 4 + x 6 )(1 + 3x 2 + 5x 4 + 7x 6 ) + (x + x 3 + x 5 + x 7 )(0 + 2x + 4x 3 + 6x 5 ) (multiplying out and using the power rule is much slower. you can use also the product rule on x(1 + x 2 + x 4 + x 6 ) 2.) c) d [sin(x) + cos(2x) tan(3x)] = d [ cos(x) 2 sin(2x) 3 sec 2 (3x) ] trig rules and chain rule. d) d (x + sin(x3 )) 15 = 15(x + sin(x 3 )) 14 (1 + d sin(x3 )) = 15(x + sin(x 3 )) 14 (1 + cos(x 3 ) d (x3 )) = 15(x + sin(x 3 )) 14 (1 + (3x 2 ) cos(x 3 )) 1 of 6
e) d ( e x 2e 2x + e 2 ex) = e x 4e 2x + e 2 ex ln(2) d (ex) = e x 4e 2x + e 2 2 ex ln(2) f) d [ln(2x) + x ln(x) + ln(sec(x))] = 1 2x 2 + (x 1 x + 1 ln(x)) + 1 (sec(x) tan(x)) sec(x) g) d x2 + x + 1 = 1 2 d x 2 + x + 1 (x2 + x + 1) = 2x + 1 2 x 2 + x + 1 2 of 6
2. (12 points) Let y = (x 2 3)(x 2 5). The derivatives are given by y = 4x(x 2)(x + 2) and y = 12(x 4/3)(x + 4/3) a) Find all critical values and determine all intervals where y is increasing. Inspection with no algebra shows the critical values are 2, 0, 2, and the increasing intervals are ( 2, 0) and (2, ). a) Find all inflection values and determine all intervals where y is concave up. The inflection values are ± 4/3. The intervals where y is concave up are (, 4/3) and ( 4/3, ). c) Make a neat sketch of y on the axis below, labeling all important features. The critical points are ( 2, 1), (0, 15), and (2, 1). The inflection points are ( 4/3, 55/9) and ( 4/3, 55/9). There are no asymptotes of any kind. 3 of 6
3. (5 points) Use the definition of the derivative to compute the derivative of f(x) = 1/x at x = 2. [You must use the definition. We know from the derivative rules that the answer will be 1/2 2 = 1/4. f f(x) f(2) (2) = = x 2 x 2 x 2 ( 2 x) 2x x 2 = ( 2 x x 2 2x(x 2) ) = x 2 ( 1 x 2 2x ) = ( 1 4 ) = 1 x 1 2 x 2 = 4. (15 pts) The function f(x) = ln(x 8 ) x in defined on the closed interval [6, 9]. Find the maximum and minimum value of f(x) on that interval. f (x) = 1 x 8 (8x7 ) 1 = 8 x 1. So the critical values f (x) = 0 occur when x = 8. f (x) is otherwise defined on the interval so there are no other critical values. f (x) = 8/x 2, which is negative so f(x) is concave down on the interval, so the critical point is the maximum value. The minimum value occurs at one of the endpoints, either 8 ln(9) 9 or 8 ln(6) 6, and you would need a calculator to determine which. 4 of 6
5. (25 pts) Find each of the following its: a) x 3 x 2 9 A polynomial is continuous everywhere, so 3 2 9 = 0. b) x 3 x 2 + 4x + 3 x 2 9 The it of the numerator is 48 and the it of the denominator is 0, so the it does not converge. Since x 2 + 9 > 0 for x > 3 and x 2 + 9 < 0 for x < 3, x 2 + 4x + 3 = and x 3 x 2 9 x 2 + 4x + 3 = + x 3 + x 2 9 So it the total it is neither one. The it does not exist. c) x 0 x ln(x + e 2 ) sin(3x) cos(x) x 0 3x ln(x + e 2 ) sin(3x) 3 cos(x) = x 0 1 ln(e2 ) 3 1 = 2/3 d) x 0 + x 2 7 x The numerator approaches 7, the denominator approaches 0 through positive values since x > 0, so the it it +. e) x x 7 7x + 4 1 + 3x 3 + 5x 5 + 7x 7 x 1 7x 6 + 4x 7 x 7 + 3x 4 + 5x 2 + 7 = 1/7 5 of 6
6. (7 pts.) Let f(x) = 9 x 3. Find the equation for the tangent line to f(x) at x = 2. f (x) = 3x 2, so the slope of the tangent line at x = 2 is f (2) = 12. The value of f at 2 is 9 2 3 = 1. The equation of the tangent line is y = 12(x 2) + 1. 7. (8 pts.) Graph the function f(x) below. Clearly label any point of discontinuity, and state what type of discontinuity it is. Clearly label any point of non-differentiability. x for 1 x f(x) = 1 for 3 x < 1 x for x < 3 The graph is the graph of the absolute value function with the section from 3 to 1 replaced by a horizontal line of height 1. There are two jump discontinuities at 1 and 3. The function f(x) is continuous at the corner of the absolute value at x = 0, however the derivative is not defined there, so the is a point of non-differentiability there. 6 of 6